What is the change in Gibbs' free energy for #"C"(s) + "O"_2(g) -> "CO"_2(g) + "393.51 kJ/mol"#?

Question

The #∆S^@# for the reaction,

#"C"(s) + "O"_2(g) -> "CO"_2(g) + "393.51 kJ/mol"#,

at #"308.62 K"# is #"0.004 kJ/mol"cdot"K"#.

Find the value of #∆G^@# at #"308.62 K"#.

What would be the answer in #"kJ/mol"#? Thanks!

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Answer 1 · 2018-03-18T04:47:30

Follow the steps below

Look at the reaction and determine the #DeltaH^@#
Since the reaction is exothermic: #DeltaH^@= -393.51 ("kJ")/("mol")#
Given #DeltaS^@ = 0.004 ("kJ")/("mol"*"K")#
Given #T= "308.62 K"#
Use the formula: #DeltaG^@=DeltaH^@-TDeltaS^@#
Plug in givens: #DeltaG^@=-393.51 ("kJ")/("mol")-(308.62 cancel("K"))*(0.004 ("kJ")/("mol"*cancel("K")))#

#= -394.74 ("kJ")/("mol") #

Answer 2 · 2018-03-18T05:08:25

#-394.74# #" kJ"/"mol"#

To solve this question, use the standard-state free energy of reaction equation:

#Delta"G"_0=DeltaH_0 - TDeltaS_0#

Given that you were given all the values for enthalpy, temperature, and entropy:

#Delta"G"_0=(-393.51 " kJ"/"mol") - (308.62" K")(0.004 "kJ"/("mol" *"K"))#

#Delta"G"_0= -394.74# #" kJ"/"mol"#

Since the value Delta #"G"_0# is negative, this reaction is thermodynamically favorable.